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Section 3.3 : Circles

6. Determine the radius and center of the following circle. If the equation is not the equation of a circle clearly explain why not.

\[{x^2} + {y^2} + 14x - 8y + 56 = 0\]

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To do this problem we need to complete the square on the \(x\) and \(y\) terms. To help with this we’ll first get the number on the right side and group the \(x\) and \(y\) terms as follows.

\[{x^2} + 14x + {y^2} - 8y = - 56\] Show Step 2

Here are the numbers we need to complete the square for both \(x\) and \(y\).

\[x:\,{\left( {\frac{{14}}{2}} \right)^2} = {\left( 7 \right)^2} = 49\hspace{0.5in}y:\, {\left( {\frac{{ - 8}}{2}} \right)^2} = {\left( { - 4} \right)^2} = 16\] Show Step 3

Now, complete the square.

\[\begin{align*}{x^2} + 14x + 49 + {y^2} - 8y + 16 & = - 56 + 49 + 16\\ {\left( {x + 7} \right)^2} + {\left( {y - 4} \right)^2} & = 9\end{align*}\]

Don’t forget to add the numbers from Step 2 to both sides of the equation!

Show Step 4

So, we have the equation in standard form and so we can quickly identify the radius and center of the circle.

\[{\mbox{Radius : }}r = 3\hspace{0.25in}\hspace{0.25in}{\mbox{Center : }}\left( { - 7,4} \right)\]

If you don’t recall how to get the radius and center from the standard form of a circle check out Problems 3 – 5 in this section for some practice.